7 Special Functions
7Special FunctionsIn this chapter we will look at some additional functions which arise often inphysical applications and are eigenfunctions for some Sturm-Liouville boundary value problem. We begin with a collection of special functions, called theclassical orthogonal polynomials. These include such polynomial functions asthe Legendre polynomials, the Hermite polynomials, the Tchebychef and theGegenbauer polynomials. Also, Bessel functions occur quite often. We willspend more time exploring the Legendre and Bessel functions. These functions are typically found as solutions of differential equations using powerseries methods in a first course in differential equations.7.1 Classical Orthogonal PolynomialsWe begin by noting that the sequence of functions {1, x, x2 , . . .} is a basis oflinearly independent functions. In fact, by the Stone-Weierstrass Approximation Theorem this set is a basis of L2σ (a, b), the space of square integrablefunctions over the interval [a, b] relative to weight σ(x). We are familiar withbeing able to expand functions over this basis, since the expansions are justpower series representations of the functions,f (x) Xcn xn .n 0However, this basis is not an orthogonal set of basis functions. One can easily see this by integrating the product of two even, or two odd, basis functionswith σ(x) 1 and (a, b) ( 1, 1). For example,Z 12 1, x2 x0 x2 dx .3 1Since we have found that orthogonal bases have been useful in determiningthe coefficients for expansions of given functions, we might ask if it is possible
2067 Special Functionsto obtain an orthogonal basis involving these powers of x. Of course, finitecombinations of these basis element are just polynomials!OK, we will ask.“Given a set of linearly independent basis vectors, canone find an orthogonal basis of the given space?” The answer is yes. Werecall from introductory linear algebra, which mostly covers finite dimensionalvector spaces, that there is a method for carrying this out called the GramSchmidt Orthogonalization Process. We will recall this process for finitedimensional vectors and then generalize to function spaces.Fig. 7.1. The basis a1 , a2 , and a3 , of R3 considered in the text.Let’s assume that we have three vectors that span R3 , given by a1 , a2 ,and a3 and shown in Figure 7.1. We seek an orthogonal basis e1 , e2 , and e3 ,beginning one vector at a time.First we take one of the original basis vectors, say a1 , and definee1 a1 .Of course, we might want to normalize our new basis vectors, so we woulddenote such a normalized vector with a “hat”:ê1 e1,e1 where e1 e1 · e1 .Next, we want to determine an e2 that is orthogonal to e1 . We take anotherelement of the original basis, a2 . In Figure 7.2 we see the orientation of thevectors. Note that the desired orthogonal vector is e2 . Note that a2 can bewritten as a sum of e2 and the projection of a2 on e1 . Denoting this projectionby pr1 a2 , we then havee2 a2 pr1 a2 .(7.1)We recall the projection of one vector onto another from our vector calculusclass.a2 · e1pr1 a2 e1 .(7.2)e21
7.1 Classical Orthogonal Polynomials207Fig. 7.2. A plot of the vectors e1 , a2 , and e2 needed to find the projection of a2 ,on e1 .Note that this is easily proven by writing the projection as a vector of lengtha2 cos θ in direction ê1 , where θ is the angle between e1 and a2 . Using thedefinition of the dot product, a · b ab cos θ, the projection formula follows.Combining Equations (7.1)-(7.2), we find thate2 a2 a2 · e1e1 .e21(7.3)It is a simple matter to verify that e2 is orthogonal to e1 :a2 · e1e1 · e1e21 a2 · e1 a2 · e1 0.e2 · e1 a2 · e1 (7.4)Now, we seek a third vector e3 that is orthogonal to both e1 and e2 . Pictorially, we can write the given vector a3 as a combination of vector projectionsalong e1 and e2 and the new vector. This is shown in Figure 7.3. Then wehave,a3 · e1a3 · e2e3 a3 e1 e2 .(7.5)2e1e22Again, it is a simple matter to compute the scalar products with e1 and e2to verify orthogonality.We can easily generalize the procedure to the N -dimensional case.Gram-Schmidt Orthogonalization in N -DimensionsLet an , n 1, ., N be a set of linearly independent vectors in RN .Then, an orthogonal basis can be found by setting e1 a1 and forn 1,n 1X an · ejen an ej .(7.6)e2jj 1
2087 Special FunctionsFig. 7.3. A plot of the vectors and their projections for determining e3 .Now, we can generalize this idea to (real) function spaces.Gram-Schmidt Orthogonalization for Function SpacesLet fn (x), n N0 {0, 1, 2, . . .}, be a linearly independent sequence of continuous functions defined for x [a, b]. Then, anorthogonal basis of functions, φn (x), n N0 can be found and isgiven byφ0 (x) f0 (x)andφn (x) fn (x) n 1Xj 0 fn , φj φj (x),kφj k2n 1, 2, . . . .(7.7)Here we are using inner products relative to weight σ(x), f, g Zbf (x)g(x)σ(x) dx.(7.8)aNote the similarity between the orthogonal basis in (7.7) and the expressionfor the finite dimensional case in Equation (7.6).Example 7.1. Apply the Gram-Schmidt Orthogonalization process to the setfn (x) xn , n N0 , when x ( 1, 1) and σ(x) 1.First, we have φ0 (x) f0 (x) 1. Note thatZ 11φ20 (x) dx .2 1We could use this result to fix the normalization of our new basis, but we willhold off on doing that for now.Now, we compute the second basis element:
7.2 Legendre Polynomials f1 , φ0 φ0 (x)kφ0 k2 x, 1 x 1 x,k1k2209φ1 (x) f1 (x) (7.9)since x, 1 is the integral of an odd function over a symmetric interval.For φ2 (x), we have f2 , φ0 f2 , φ1 φ0 (x) φ1 (x)2kφ0 kkφ1 k2 x2 , 1 x2 , x x2 1 xk1k2kxk2R1 2x dx2 x 1R1 1 dx1 x2 .3φ2 (x) f2 (x) (7.10)So far, we have the orthogonal set {1, x, x2 31 }. If one chooses to normalize these by forcing φn (1) 1, then one obtains the classical Legendrepolynomials, Pn (x) φ1 (x). Thus,P2 (x) 1(3x2 1).2Note that this normalization is different than the usual one. In fact, we seethat P2 (x) does not have a unit norm,Z 12kP2 k2 P22 (x) dx .5 1The set of Legendre polynomials is just one set of classical orthogonalpolynomials that can be obtained in this way. Many had originally appearedas solutions of important boundary value problems in physics. They all havesimilar properties and we will just elaborate some of these for the Legendrefunctions in the next section. Other orthogonal polynomials in this group areshown in Table 7.1.For reference, we also note the differential equations satisfied by thesefunctions.7.2 Legendre PolynomialsIn the last section we saw the Legendre polynomials in the context of orthogonal bases for a set of square integrable functions in L2 ( 1, 1). In yourfirst course in differential equations, you saw these polynomials as one of thesolutions of the differential equation
2107 Special erTchebychef of the 1st kindTchebychef of the 2nd kindJacobiSymbol Intervalσ(x)2Hn (x) ( , )e xα xLn (x)[0, )ePn (x)(-1,1)1Cnλ (x)(-1,1)(1 x2 )λ 1/2Tn (x)(-1,1)(1 x2 ) 1/2Un (x)(-1,1)(1 x2 ) 1/2(ν,µ)Pn (x) (-1,1) (1 x)ν (1 x)µTable 7.1. Common classical orthogonal polynomials with the interval and weightfunction used to define them.PolynomialDifferential EquationHermitey ′′ 2xy ′ 2ny 0′′Laguerrexy (α 1 x)y ′ ny 0Legendre(1 x2 )y ′′ 2xy ′ n(n 1)y 0Gegenbauer(1 x2 )y ′′ (2n 3)xy ′ λy 0Tchebychef of the 1st kind(1 x2 )y ′′ xy ′ n2 y 02 ′′Jacobi(1 x )y (ν µ (µ ν 2)x)y ′ n(n 1 µ ν)y 0Table 7.2. Differential equations satisfied by some of the common classical orthogonal polynomials.(1 x2 )y ′′ 2xy ′ n(n 1)y 0,n N0 .(7.11)Recall that these were obtained by using power series expansion methods. Inthis section we will explore a few of the properties of these functions.For completeness, we recall the solution of Equation (7.11) using the powerseries method. We assume that the solution takes the formy(x) Xak xk .k 0The goal is to determine the coefficients, ak . Inserting this series into Equation(7.11), we have(1 x2 )or Xk 2 Xk 0k(k 1)ak xk 2 k(k 1)ak xk 2 Xk 2 X2ak kxk k 0k(k 1)ak xk Xn(n 1)ak xk 0,k 0 X[ 2k n(n 1)] ak xk 0.k 0We can combine some of these terms: Xk 2k(k 1)ak xk 2 Xk 0[ k(k 1) 2k n(n 1)] ak xk 0.
7.2 Legendre Polynomials211Further simplification yields Xk 2k(k 1)ak xk 2 Xk 0[n(n 1) k(k 1)] ak xk 0.We need to collect like powers of x. This can be done by reindexing each sum.In the first sum, we let m k 2, or k m 2. In the second sum weindependently let k m. Then all powers of x are of the form xm . This gives Xm(m 2)(m 1)am 2 x m 0 Xm 0[n(n 1) m(m 1)] am xm 0.Combining these sums, we have Xm 0[(m 2)(m 1)am 2 (n(n 1) m(m 1))am ] xm 0.This has to hold for all x. So, the coefficients of xm must vanish:(m 2)(m 1)am 2 (n(n 1) m(m 1))am .Solving for am 2 , we obtain the recursion relationam 2 n(n 1) m(m 1)am ,(m 2)(m 1)m 0.Thus, am 2 is proportional to am . We can iterate and show that each coefficient is either proportional to a0 or a1 . However, for n an integer, sooner, orlater, m n and the series truncates. am 0 for m n. Thus, we obtainpolynomial solutions. These polynomial solutions are the Legendre polynomials, which we designate as y(x) Pn (x). Furthermore, for n an even integer,Pn (x) is an even function and for n an odd integer, Pn (x) is an odd function.Actually, this is a trimmed down version of the method. We would need tofind a second linearly independent solution. We will not discuss these solutionsand leave that for the interested reader to investigate.7.2.1 The Rodrigues FormulaThe first property that the Legendre polynomials have is the Rodrigues formula:1 dn 2Pn (x) n(x 1)n , n N0 .(7.12)2 n! dxnFrom the Rodrigues formula, one can show that Pn (x) is an nth degree polynomial. Also, for n odd, the polynomial is an odd function and for n even, thepolynomial is an even function.As an example, we determine P2 (x) from Rodrigues formula:
2127 Special FunctionsP2 (x) 1 d2 2(x 1)222 2! dx21 d2 4(x 2x2 1)8 dx21 d(4x3 4x)8 dx1(12x2 4)81(3x2 1).2(7.13)Note that we get the same result as we found in the last section using orthogonalization.One can systematically generate the Legendre polynomials in tabular formas shown in Table 7.2.1. In Figure 7.4 we show a few Legendre polynomials.ndn(x2 1)n(x2 1)ndxn0111x2 12x2x4 2x2 112x2 46423 x 3x 3x 1 120x3 72x12n n!11218148Pn (x)1x1(3x2 1)21(5x3 3x)2Table 7.3. Tabular computation of the Legendre polynomials using the .20.40.60.81x–0.5–1Fig. 7.4. Plots of the Legendre polynomials P2 (x), P3 (x), P4 (x), and P5 (x).
7.2 Legendre Polynomials2137.2.2 Three Term Recursion FormulaThe classical orthogonal polynomials also satisfy three term recursion formulae. In the case of the Legendre polynomials, we have(2n 1)xPn (x) (n 1)Pn 1 (x) nPn 1 (x),n 1, 2, . . . .(7.14)This can also be rewritten by replacing n with n 1 as(2n 1)xPn 1 (x) nPn (x) (n 1)Pn 2 (x),n 1, 2, . . . .(7.15)We will prove this recursion formula in two ways. First we use the orthogonality properties of Legendre polynomials and the following lemma.Lemma 7.2. The leading coefficient of xn in Pn (x) is1 (2n)!2n n! n! .Proof. We can prove this using Rodrigues formula. first, we focus on theleading coefficient of (x2 1)n , which is x2n . The first derivative of x2n is2nx2n 1 . The second derivative is 2n(2n 1)x2n 2 . The jth derivative isdj x2n [2n(2n 1) . . . (2n j 1)]x2n j .dxjThus, the nth derivative is given bydn x2n [2n(2n 1) . . . (n 1)]xn .dxnThis proves that Pn (x) has degree n. The leading coefficient of Pn (x) can nowbe written as11n(n 1) . . . 1[2n(2n 1) . . . (n 1)] n [2n(2n 1) . . . (n 1)]2n n!2 n!n(n 1) . . . 11 (2n)! n.(7.16)2 n! n!In order to prove the three term recursion formula we consider the expression (2n 1)xPn 1 (x) nPn (x). While each term is a polynomial of degreen, the leading order terms cancel. We need only look at the coefficient of theleading order term first expression. It is(2n 1)12n 1 (n(2n 2)!1(2n 1)!(2n 1)! n 1 . 1)! (n 1)!2(n 1)! (n 1)!2n 1 [(n 1)!]2The coefficient of the leading term for nPn (x) can be written as 1 (2n)!2n1(2n 1)!(2n 1)!n n n.2 n! n!2n22n 1 (n 1)! (n 1)! 2n 1 [(n 1)!]2
2147 Special FunctionsIt is easy to see that the leading order terms in (2n 1)xPn 1 (x) nPn (x)cancel.The next terms will be of degree n 2. This is because the Pn ’s are eithereven or odd functions, thus only containing even, or odd, powers of x. Weconclude that(2n 1)xPn 1 (x) nPn (x) polynomial of degree n 2.Therefore, since the Legendre polynomials form a basis, we can write thispolynomial as a linear combination of of Legendre polynomials:(2n 1)xPn 1 (x) nPn (x) c0 P0 (x) c1 P1 (x) . . . cn 2 Pn 2 (x). (7.17)Multiplying Equation (7.17) by Pm (x) for m 0, 1, . . . , n 3, integratingfrom 1 to 1, and using orthogonality, we obtain0 cm kPm k2 ,m 0, 1, . . . , n 3.R1[Note: 1 xk Pn (x) dx 0 for k n 1. Thus, 1 xPn 1 (x)Pm (x) dx 0for m n 3.]Thus, all of these cm ’s are zero, leaving Equation (7.17) asR1(2n 1)xPn 1 (x) nPn (x) cn 2 Pn 2 (x).The final coefficient can be found by using the normalization condition,Pn (1) 1. Thus, cn 2 (2n 1) n n 1.7.2.3 The Generating FunctionA second proof of the three term recursion formula can be obtained from thegenerating function of the Legendre polynomials. Many special functions havesuch generating functions. In this case it is given by X1 Pn (x)tn ,g(x, t) 1 2xt t2n 0 x 1, t 1.(7.18)This generating function occurs often in applications. In particular, itarises in potential theory, such as electromagnetic or gravitational potentials.These potential functions are 1r type functions. For example, the gravitationalpotential between the Earth and the moon is proportional to the reciprocalof the magnitude of the difference between their positions relative to somecoordinate system. An even better example, would be to place the origin atthe center of the Earth and consider the forces on the non-pointlike Earth dueto the moon. Consider a piece of the Earth at position r1 and the moon atposition r2 as shown in Figure 7.5. The tidal potential Φ is proportional toΦ 111 p p 2, r2 r1 (r2 r1 ) · (r2 r1 )r1 2r1 r2 cos θ r22
7.2 Legendre Polynomials215Fig. 7.5. The position vectors used to describe the tidal force on the Earth due tothe moon.where θ is the angle between r1 and r2 .Typically, one of the position vectors is much larger than the other. Let’sassume that r1 r2 . Then, one can write111rΦ p 2 2 .2r2r1 2r1 r2 cos θ r2r11 2 r2 cos θ rr12Now, define x cos θ and t rr12 . We then have the tidal potential is proportional to the generating function for the Legendre polynomials! So, we canwrite the tidal potential as n r11 XΦ Pn (cos θ).r2 n 0r2The first term in the expansion is the gravitational potential that gives theusual force between the Earth and the moon. [Recall that the force is thegradient of the potential, F 1r .] The next terms will give expressionsfor the tidal effects.Now that we have some idea as to where this generating function mighthave originated, we can proceed to use it. First of all, the generating functioncan be used to obtain special values of the Legendre polynomials.Example 7.3. Evaluate Pn (0). Pn (0) is found by considering g(0, t). Settingx 0 in Equation (7.18), we have X1g(0, t) Pn (0)tn .1 t2n 0(7.19)We can use the binomial expansion to find our final answer. [See the lastsection of this chapter for a review.] Namely, we have113 1 t2 t4 . . . .2281 tComparing these expansions, we have the Pn (0) 0 for n odd and for evenintegers one can show (see Problem 7.10) that
2167 Special FunctionsP2n (0) ( 1)n(2n 1)!!,(2n)!!(7.20)where n!! is the double factorial, n(n 2) . . . (3)1, n 0, odd,n!! n(n 2) . . . (4)2, n 0, even, . 1n 0, 1Example 7.4. Evaluate Pn ( 1). This is a simpler problem. In this case we have11 1 t t2 t3 . . . .g( 1, t) 1 t1 2t t2Therefore, Pn ( 1) ( 1)n .We can also use the generating function to find recursion relations. Toprove the three term recursion (7.14) that we introduced above, then we needonly differentiate the generating function with respect to t in Equation (7.18)and rearrange the result. First note that gx tx t g(x, t).23/2 t1 2xt t2(1 2xt t )Combining this with X g nPn (x)tn 1 , tn 0we have(x t)g(x, t) (1 2xt t2 ) XnPn (x)tn 1 .n 0Inserting the series expression for g(x, t) and distributing the sum on the rightside, we obtain(x t) XPn (x)tn n 0 Xn 0 XnPn (x)tn 1 2nxPn (x)tn n 0 XnPn (x)tn 1 .n 0Rearranging leads to three separate sums: Xn 0nPn (x)tn 1 X(2n 1)xPn (x)tn n 0 X(n 1)Pn (x)tn 1 0. (7.21)n 0Each term contains powers of t that we would like to combine into a singlesum. This is done by reindexing. For the first sum, we could use the new indexk n 1. Then, the first sum can be written Xn 0nPn (x)tn 1 X(k 1)Pk 1 (x)tk .k 1
7.2 Legendre Polynomials217Using different indices is just another way of writing out the terms. Note that XnPn (x)tn 1 0 P1 (x) 2P2 (x)t 3P3 (x)t2 . . .n 0and X(k 1)Pk 1 (x)tk 0 P1 (x) 2P2 (x)t 3P3 (x)t2 . . .k 1actually give the same sum. The indices are sometimes referred to as dummyindices because they do not show up in the expanded expression and can bereplaced with another letter.If we want to do so, we could now replace all of the k’s with n’s. However,we will leave the k’s in the first term and now reindex the next sums inEquation (7.21). The second sum just needs the replacement n k and thelast sum we reindex using k n 1. Therefore, Equation (7.21) becomes X(k 1)Pk 1 (x)tk k 1 X(2k 1)xPk (x)tk k 0 XkPk 1 (x)tk 0. (7.22)k 1We can now combine all of the terms, noting the k 1 term is automaticallyzero and the k 0 terms giveP1 (x) xP0 (x) 0.(7.23)Of course, we know this already. So, that leaves the k 0 terms: Xk 1[(k 1)Pk 1 (x) (2k 1)xPk (x) kPk 1 (x)] tk 0.(7.24)Since this is true for all t, the coefficients of the tk ’s are zero, or(k 1)Pk 1 (x) (2k 1)xPk (x) kPk 1 (x) 0,k 1, 2, . . . .There are other recursion relations. For example,′′Pn 1(x) Pn 1(x) (2n 1)Pn (x).(7.25)This can be proven using the generating function by differentiating g(x, t)with respect to x and rearranging the resulting infinite series just as in thislast manipulation. This will be left as Problem 7.4.Another use of the generating function is to obtain the normalization constant. Namely, kPn k2 . Squaring the generating function, we have" #2 X XX1n Pn (x)t Pn (x)Pm (x)tn m .(7.26)1 2xt t2n 0n 0 m 0
2187 Special FunctionsIntegrating from -1 to 1 and using the orthogonality of the Legendre polynomials, we haveZ1 1Z 1 X Xdxn m tPn (x)Pm (x) dx1 2xt t2 1n 0 m 0Z 1 X t2nPn2 (x) dx.However, one can show thatZ 1 1(7.27) 1n 0dx1 ln1 2xt t2t 1 t1 t .Expanding this expression about t 0, we obtain X 11 t2ln t2n .t1 t2n 1n 0Comparing this result with Equation (7.27), we find thatZ 12kPn k2 Pn (x)Pm (x) dx .2n 1 1(7.28)7.2.4 Eigenfunction ExpansionsFinally, we can expand other functions in this orthogonal basis. This is justa generalized Fourier series. A Fourier-Legendre series expansion for f (x) on[ 1, 1] takes the form Xf (x) cn Pn (x).(7.29)n 0As before, we can determine the coefficients by multiplying both sides byPm (x) and integrating. Orthogonality gives the usual form for the generalizedFourier coefficients. In this case, we havecn where f, Pn f, Pn ,kPn k2Z1f (x)Pn (x) dx. 12We have just found kPn k2 2n 1. Therefore, the Fourier-Legendre coefficients areZ2n 1 1cn f (x)Pn (x) dx.(7.30)2 1
7.2 Legendre PolynomialsExample 7.5. Expand f (x) x3 in a Fourier-Legendre series.We simply need to computeZ2n 1 1 3cn x Pn (x) dx.2 1219(7.31)We first note thatZ1xm Pn (x) dx 0for m n. 1This is simply proven using Rodrigues formula. Inserting Equation (7.12), wehaveZ 1Z 11dnxm n (x2 1)n dx.xm Pn (x) dx n2 n! 1dx 1Since m n, we can integrate by parts m-times to show the result, usingPn (1) 1 and Pn ( 1) ( 1)n . As a result, we will have for this examplethat cn 0 for n 3.R1We could just compute 1 x3 Pm (x) dx for m 0, 1, 2, . . . outright. But,noting that x3 is an odd function, we easily confirm that c0 0 and c2 0.This leaves us with only two coefficients to compute. These are3c1 2and7c3 2Z13x 1Z 1 1x4 dx 35 123(5x 3x) dx .25Thus,32P1 (x) P3 (x).55Of course, this is simple to check using Table 7.2.1: 3232 13P1 (x) P3 (x) x (5x 3x) x3 .5555 2x3 Well, maybe we could have guessed this without doing any integration. Let’ssee,1x3 c1 x c2 (5x3 3x)235 (c1 c2 )x c2 x3 .22Equating coefficients of like terms, we have that c2 (7.32)25and c1 23 c2 35 .
2207 Special FunctionsExample 7.6. Expand the Heaviside function in a Fourier-Legendre series.The Heaviside function is defined as 1, x 0,H(x) (7.33)0, x 0.In this case, we cannot find the expansion coefficients without some integration. We have to computeZ2n 1 1cn f (x)Pn (x) dx2 1Z2n 1 1 Pn (x) dx, n 0, 1, 2, . . . .(7.34)20For n 0, we haveZ1 1dx 2 0For n 1, we make use of the identity (7.25)Z1 1 ′′[P(x) Pn 1(x)] dx cn 2 0 n 1c0 1.2to find1[Pn 1 (0) Pn 1 (0)].2Thus, the Fourier-Bessel series for the Heaviside function isf (x) 1 1X [Pn 1 (0) Pn 1 (0)]Pn (x).2 2 n 1We need to evaluate Pn 1 (0) Pn 1 (0). Since Pn (0) 0 for n odd, thecn ’s vanish for n even. Letting n 2k 1, we havef (x) 1 1X[P2k 2 (0) P2k (0)]P2k 1 (x). 2 2k 1We can use Equation (7.20),P2k (0) ( 1)k(2k 1)!!,(2k)!!to compute the coefficients: 1 1X [P2k 2 (0) P2k (0)]P2k 1 (x)2 2k 1 1 1Xk 1 (2k 3)!!k (2k 1)!! ( 1) ( 1)P2k 1 (x)2 2(2k 2)!!(2k)!!k 1 1 1X(2k 3)!!2k 1 ( 1)k1 P2k 1 (x)2 2(2k 2)!!2kf (x) 1 1 2 2k 1 Xk 1( 1)k(2k 3)!! 4k 1P2k 1 (x).(2k 2)!! 2k(7.35)
7.3 Gamma Function221The sum of the first 21 terms are shown in Figure 7.6. We note the slow convergence to the Heaviside function. Also, we see that the Gibbs phenomenonis present due to the jump discontinuity at x 0.Partial Sum of Fourier-Legendre .60.8xFig. 7.6. Sum of first 21 terms for Fourier-Legendre series expansion of Heavisidefunction.7.3 Gamma FunctionAnother function that often occurs in the study of special functions is theGamma function. We will need the Gamma function in the next section onBessel functions.For x 0 we define the Gamma function asZ Γ (x) tx 1 e t dt, x 0.(7.36)0The Gamma function is a generalization of the factorial function. In fact,we haveΓ (1) 1andΓ (x 1) xΓ (x).The reader can prove this identity by simply performing an integration byparts. (See Problem 7.7.) In particular, for integers n Z , we then haveΓ (n 1) nΓ (n) n(n 1)Γ (n 2) n(n 1) · · · 2Γ (1) n!.We can also define the Gamma function for negative, non-integer valuesof x. We first note that by iteration on n Z , we have
2227 Special FunctionsΓ (x n) (x n 1) · · · (x 1)xΓ (x),x 0,x n 0.Solving for Γ (x), we then findΓ (x) Γ (x n),(x n 1) · · · (x 1)x n x 0Note that the Gamma function is undefined at zero and the negative integers.Example 7.7. We now prove thatΓ 1 π.2This is done by direct computation of the integral: Z 11Γ t 2 e t dt.20Letting t z 2 , we haveΓ Z 21 2e z dz.20Due to the symmetry of the integrand, we obtain the classic integral Z 21Γ e z dz,2 which can be performed using a standard trick. Consider the integralZ 2I e x dx. Then,I2 Z 2e x dx Z 2e y dy. Note that we changed the integration variable. This will allow us to write thisproduct of integrals as a double integral:Z Z 222I e (x y ) dxdy. This is an integral over the entire xy-plane. We can transform this Cartesianintegration to an integration over polar coordinates. The integral becomesZ 2π Z 2I2 e r rdrdθ.00This is simple to integrate and we have I 2 π. So, the final result is foundby taking the square root of both sides: 1Γ I π.2
7.4 Bessel Functions223We have seen that the factorial function can be written in terms of Gammafunctions. One can write the even and odd double factorials as(2n)!! 2n n!,(2n 1)!.2n n!(2n 1)!! In particular, one can write1(2n 1)!! Γ (n ) π.22nAnother useful relation, which we only state, isΓ (x)Γ (1 x) π.sin πx7.4 Bessel FunctionsAnother important differential equation that arises in many physics applications isx2 y ′′ xy ′ (x2 p2 )y 0.(7.37)This equation is readily put into self-adjoint form as(xy ′ )′ (x p2)y 0.x(7.38)This equation was solved in the first course on differential equations usingpower series methods, namely by using the Frobenius Method. One assumesa series solution of the formy(x) Xan xn s ,n 0and one seeks allowed values of the constant s and a recursion relation for thecoefficients, an . One finds that s p andan an 2,(n s)2 p2n 2.One solution of the differential equation is the Bessel function of the firstkind of order p, given asy(x) Jp (x) X x 2n p( 1)n.Γ (n 1)Γ (n p 1) 2n 0(7.39)In Figure 7.7 we display the first few Bessel functions of the first kind of integer order. Note that these functions can be described as decaying oscillatoryfunctions.
2247 Special –0.2–0.4Fig. 7.7. Plots of the Bessel functions J0 (x), J1 (x), J2 (x), and J3 (x).A second linearly independent solution is obtained for p not an integer asJ p (x). However, for p an integer, the Γ (n p 1) factor leads to evaluationsof the Gamma function at zero, or negative integers, when p is negative. Thus,the above series is not defined in these cases.Another method for obtaining a second linearly independent solution isthrough a linear combination of Jp (x) and J p (x) asNp (x) Yp (x) cos πpJp (x) J p (x).sin πp(7.40)These functions are called the Neumann functions, or Bessel functions of thesecond kind of order p.In Figure 7.8 we display the first few Bessel functions of the second kind ofinteger order. Note that these functions are also decaying oscillatory functions.However, they are singular at x 0.In many applications these functions do not satisfy the boundary condition that one desires a bounded solution at x 0. For example, one standardproblem is to describe the oscillations of a circular drumhead. For this problem one solves the wave equation using separation of variables in cylindricalcoordinates. The r equation leads to a Bessel equation. The Bessel functionsolutions describe the radial part of the solution and one does not expect asingular solution at the center of the drum. The amplitude of the oscillationmust remain finite. Thus, only Bessel functions of the first kind can be used.Bessel functions satisfy a variety of properties, which we will only list atthis time for Bessel functions of the first kind.Derivative Identitiesd p[x Jp (x)] xp Jp 1 (x).dx(7.41)
7.4 Bessel 10x–0.2–0.4–0.6–0.8–1Fig. 7.8. Plots of the Neumann functions N0 (x), N1 (x), N2 (x), and N3 (x). d px Jp (x) x p Jp 1 (x).dx(7.42)Recursion FormulaeJp 1 (x) Jp 1 (x) 2pJp (x).xJp 1 (x) Jp 1 (x) 2Jp′ (x).OrthogonalityZ axa2x2[Jp 1 (jpn )] δn,mxJp (jpn )Jp (jpm ) dx aa20(7.43)(7.44)(7.45)where jpn is the nth root of Jp (x), Jp (jpn ) 0, n 1, 2, . . . . A list ofsome of these roots are provided in Table 7.4.n123456789p 94p 047p 571p 050p .512p .983Table 7.4. The zeros of Bessel Functions
2267 Special FunctionsGenerating Function1ex(t t )/2 XJn (x)tn ,n Integral RepresentationZ1 πJn (x) cos(x sin θ nθ) dθ,π 0x 0, t 6 0.x 0, n Z.(7.46)(7.47)Fourier-Bessel SeriesSince the Bessel functions are an orthogonal set of eigenfunctions of aSturm-Liouville problem, we can expand square integrable functions inthis basis. In fact, the eigenvalue problem is given in the formx2 y ′′ xy ′ (λx2 p2 )y 0.(7.48) The solutions are then of the form Jp ( λx), as can be shown by makingthe substitution t λx in the differential equation.Furthermore, one can solve the differential equation on a finite domain,[0, a], with the boundary conditions: y(x) is bounded at x 0 and y(a) 0. One can show that Jp (jpn xa ) is a basis of eigenfunctions and the resultingFourier-Bessel series expansion of f (x) defined on x [0, a] isf (x) Xxcn Jp (jpn ),an 1(7.49)where the Fourier-Bessel coefficients are found using the orthogonalityrelation asZ a2xcn xf (x)Jp (jpn ) dx.(7.50)2aa2 [Jp 1 (jpn )] 0Example 7.8. Expand f (x) 1 for 0 x 1 in a Fourier-Bessel series ofthe form Xf (x) cn J0 (j0n x)n 1.We need only compute the Fourier-Bessel coefficients in Equation (7.50):cn 2[J1 (j0n )]2From Equation (7.41) we haveZ01xJ0 (j0n x) dx.(7.51)
7.5 Hypergeometric FunctionsZ1xJ0 (j0n x) dx 012j0n1Zj0nyJ0 (y) dy0Zj0nd[yJ1 (y)] dy2j0ndy01 2 [yJ1 (y)]j00nj0n1 J1 (j0n ).j0n 227(7.52)As a result, we have found that the desired Fourier-Bessel expansion is1 2 XJ0 (j0n x),j J (j )n 1 0n 1 0n0 x 1.(7.53)In Figure 7.9 we show the partial sum for the first fifty terms of this series.We see that there is slow convergence due to the Gibbs’ phenomenon
Special Functions In this chapter we will look at some additional functions which arise often in physical applications and are eigenfunctions for some Sturm-Liouville bound-ary value problem. We begin with a collection of special functions, called the . since x,1 is the integral of an
Class- VI-CBSE-Mathematics Knowing Our Numbers Practice more on Knowing Our Numbers Page - 4 www.embibe.com Total tickets sold ̅ ̅ ̅̅̅7̅̅,707̅̅̅̅̅ ̅ Therefore, 7,707 tickets were sold on all the four days. 2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches.
Aug 13, 2020 · exponential functions. Unit 5.1 –Exponential Functions & Their Graphs So far, this text has dealt mainly with algebraic functions, which include polynomial functions and rational functions. In this chapter, you will study two types of nonalgebraic functions –exponential funct
Unit 1 - Chapter 2 Oracle Built in Functions There are two types of functions in Oracle. 1) Single Row Functions: Single row or Scalar functions return a value for every row that is processed in a query. 2) Group Functions: These functions group the rows of
Functions are useful while writing SQL queries also. Functions can be applied to work on single or multiple records (rows) of a table. Depending on their application in one or multiple rows, SQL functions are categorised as Single row functions and Aggregate functions. 1.2.1 Single Row Functions These are al
29 Functions and their Graphs The concept of a function was introduced and studied in Section 7 of these notes. In this section we explore the graphs of functions. Of particular in-terest, we consider the graphs of linear functions, quadratic functions, cubic functions, square root functions,
These are functions that are not linear. Their graph is not a straight line. The degree of these functions is not 1. Non-linear functions can be: Quadratic functions – a polynomial of degree 2 Cubic functions – a polynomial of degree 3 Other higher order functions Exponential functions
Lesson 7-7B Comparing Functions Lesson Comparing Functions Chapter 7 7-7B BIG IDEA Different descriptions of functions make it possible to compare functions in a variety of ways. You have seen functions described verbally, in tables, by equations, and using graphs. With any of these descriptions, you can compare functions. Example 1
Physics 212 2010, Electricity and Magnetism Special Functions: Legendre functions, Spherical Harmonics, and Bessel Functions Note that the first
